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Thread: n-th root of natural number

  1. #1
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    n-th root of natural number

    I would like to prove that:

    If $m,n \in \mathbb{N}$ then $\sqrt[m]{n}$ is either an integer or irrational.

    Here is my attempt:

    Suppose that $\sqrt[m]{n} \in \mathbb{Q}$. Then $\exists \; p,q \in \mathbb{Z}$ such that
    $$ \sqrt[m]{n} = \frac{p}{q}$$
    which implies that $$ n = \frac{p^m}{q^m}.$$
    Now if $q \neq 1$ then $q^m | p^m$ since $n \in \mathbb{N}$.

    Because $\sqrt[m]{n} \in \mathbb{Q}$ then we can assume that $gcd(p,q)=1.$

    By the fundamental theorem of arithmetic we have that $gcd(p,q)=gcd(p^m,q^m)$ hence
    $$ gcd(p^m,q^m)=1 \implies q^m \nmid p^m \implies q =1$$
    so $n=p^m \implies \sqrt[m]{n} =p \in \mathbb{Z}$ which contradicts the fact that $\sqrt[m]{n} \in \mathbb{Q}$.

    Therefore $\forall \; m,n \in \mathbb{N}$ if $\sqrt[m]{n}$ is not an integer then it is not rational.
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  2. #2
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    Re: n-th root of natural number

    Yep, that's the way I've always approached this. In essence p and q can't share any common factors, which leads to p^m can't equal nq^m unless q = 1.
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  3. #3
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    Re: n-th root of natural number

    Don't take what follows as harsh criticism. I hope you find it constructive. You're in the right ballpark, but these things take practice to get right (as does everything).

    Quote Originally Posted by Incongruent View Post
    Suppose that $\displaystyle \sqrt[m]{n} \in \mathbb{Q}$. Then $\displaystyle \exists \; p,q \in \mathbb{Z}$ such that

    $\displaystyle \sqrt[m]{n} = \frac{p}{q}$
    If the well-known fact that your $\displaystyle p$ and $\displaystyle q$ can be chosen to be relatively prime has any bearing whatsoever on your later intended uses for them - and in your case it most certainly does - then there's no reason to not say that you've chosen them to be relatively prime right from the start. This is one of those rare recommendations that I'd wager has not a single exception in the whole of all well-written math proofs ever... that since you can choose them to be relatively prime, you should do it right up front when you first define them. The only exception is for proofs where their relative-primeness will never be mentioned or used (and that's not your case in this problem... it will be used).

    For a general rational number, you can also always choose one of them to be positive... the denominator is best/most natural. In this case, after you point out that your rational number is positive (since n is positive), you could also declare that the numerator is positive. When working divisibility problems, signs (UNITS!) aren't going to be a big deal, but it's still best to keep everything clear and correct. For example, if you don't say anything about the sign of q, then when you want to eventually declare that q^m divides 1 implies that q = 1, it's not really a justified conclusion (the justified conclusion is that q = 1 or q = -1).

    To be clear, although specifying the the signs is good (and correct), that's not nearly as important a detail as declaring upfront that you've chosen your integers to be relatively prime (since the latter is what's crucially essential to your argument).

    Maybe write:
    "Suppose that $\displaystyle \sqrt[m]{n} \in \mathbb{Q}$ where $\displaystyle m, n \in \mathbb{N}$. Then $\displaystyle \ \exists \ p,q \in \mathbb{Z}$ such that $\displaystyle p>0, q > 0, gcd(p, q) = 1,$ and $\displaystyle \ \sqrt[m]{n} = \frac{p}{q}$."

    Quote Originally Posted by Incongruent View Post
    Now if $\displaystyle q \neq 1$ then $\displaystyle q^m | p^m$ since $\displaystyle n \in \mathbb{N}$.
    True, but $\displaystyle q^m | p^m$ at this step regardless of whether or not $\displaystyle q = 1$. The $\displaystyle q \ne 1$ is a superfluous condition - one that's also apparently irrelevant to your argument.

    Write: $\displaystyle \frac{p^m}{q^m} = n \in \mathbb{Z} \Rightarrow q^m | p^m$ (It's essentially the definition of "divides").

    Quote Originally Posted by Incongruent View Post
    Because $\displaystyle \sqrt[m]{n} \in \mathbb{Q}$ then we can assume that $\displaystyle gcd(p,q)=1.$
    Again, this should've been done the moment you introduced $\displaystyle p$ and $\displaystyle q$.

    Quote Originally Posted by Incongruent View Post
    By the fundamental theorem of arithmetic we have that $\displaystyle gcd(p,q)=gcd(p^m,q^m)$ hence
    Since by this point you'd argued that gcd(p,q)=1, that's technically correct. But the way you've written it is confused to the point of being invalid. It reads as if you're using a false general theorem, and then, without mentioning the truly salient detail, applying it to your particular case of $\displaystyle gcd(p,q) = 1$ to conclude that $\displaystyle gcd(p^m,q^m) = 1$, and then not even stating that conclusion, but burying it within $\displaystyle gcd(p,q) = gcd(p^m,q^m)$.

    The justification for $\displaystyle gcd(p^m,q^m) = 1$ is NOT your "By the fundamental theorem of arithmetic we have that $\displaystyle gcd(p,q)=gcd(p^m,q^m)$", but maybe instead: "It's clear that $\displaystyle gcd(p,q) = 1 \Rightarrow gcd(p^m,q^m) = 1$".

    The statement $\displaystyle gcd(p,q)=gcd(p^m,q^m)$ isn't correct in general, but ONLY holds when $\displaystyle gcd(p,q)=1$ (ignoring signs and the pq = 0 case).

    The correct general statement would be: $\displaystyle gcd(x, y)^n=gcd(x^n, y^n) \ \forall \ x, y \in \mathbb{N}$.

    Also, although that correct general formula is both intuitively true and routinely straightforward to prove, my opinion is that it's not so trivial as to be cited as a fact without giving a more detailed justification for it. If you're writing for a math journal, then fine - skip it as "trivial", but for a class, my opinion is that it should have a more detailed and clear justification than "By the fundamental theorem of arithmetic".

    Proofs require judgment calls for how many trivial statements should be explicitly justified. What is truly trivial and so beneath investing time to talk about, and what isn't?

    Although $\displaystyle gcd(x, y)^n=gcd(x^n, y^n) \ \forall \ x, y \in \mathbb{N}$ is certainly not particularly taxing to "see", nor to prove, if you do try to prove it, you'll see why I think it demands a fuller explanation than my "It's clear that $\displaystyle gcd(p,q) = 1 \Rightarrow gcd(p^m,q^m) = 1$".

    With "It's clear that $\displaystyle gcd(p,q) = 1 \Rightarrow gcd(p^m,q^m) = 1$", anyone sufficiently competent to read your work at all will almost instantly see both why it's true and how to prove it; the speed which with they write/speak the proof essentially equals the speed at which they can write/speak. That's not the case with "$\displaystyle gcd(x, y)^n=gcd(x^n, y^n) \ \forall \ x, y \in \mathbb{N}$."

    (For $\displaystyle gcd(p,q) = 1 \Rightarrow gcd(p^m,q^m) = 1$ :: "If $\displaystyle gcd(p^m,q^m) \ne 1$, then some prime divides both $\displaystyle p^m$ and $\displaystyle q^m$, so it divides both $\displaystyle p$ and $\displaystyle q$, so it divides $\displaystyle gcd(p,q) = 1$. Contradiction.")

    Quote Originally Posted by Incongruent View Post
    $\displaystyle gcd(p^m,q^m)=1 \implies q^m \nmid p^m \implies q =1$
    This is just incorrect, because $\displaystyle q^m | p^m$ is in fact true. Also, your "implies" claims there don't make sense to me.
    I think what you're going for is this:

    "Since $\displaystyle q^m | p^m$, have that $\displaystyle q^m | gcd(p^m,q^m)$. But $\displaystyle gcd(p^m,q^m) = 1$, so $\displaystyle q^m | 1$, and so $\displaystyle q = 1$."

    Quote Originally Posted by Incongruent View Post
    so $\displaystyle n=p^m \implies \sqrt[m]{n} =p \in \mathbb{Z}$ which contradicts the fact that $\displaystyle \sqrt[m]{n} \in \mathbb{Q}$.
    No, it doesn't contradict it. Every integer is also a rational number. Try to clarify in your own mind exactly what you're assuming, concluding, and meaning as you go.
    Reread your own argument. What was your initial assumption about that $\displaystyle \sqrt[m]{n}$? What have you deduced from that assumption?

    What you've actually shown at this step (had you done everything correctly) is: IF $\displaystyle \sqrt[m]{n}$ is rational, THEN it must be an integer.

    But that's all you need to show to have completed proving the theorem, IF you can wrap it all up correctly... word your final conclusion correctly.
    Last edited by johnsomeone; Jun 8th 2015 at 08:32 AM.
    Thanks from Wander, Incongruent and sakonpure6
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  4. #4
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    Re: n-th root of natural number

    Thank you. I will look at this in detail as soon as I can. I just wanted to say thanks for everything you wrote. I really don't feel confident proving this stuff rigorously and everything you have pointed out has help a lot.
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  5. #5
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    Re: n-th root of natural number

    That's a good clear post by johnsomeone but I would argue against this:
    Quote Originally Posted by johnsomeone View Post
    "It's clear that $\displaystyle gcd(p,q) = 1 \Rightarrow gcd(p^m,q^m) = 1$".
    Sentences that claim that something is clear or obvious are bad mathematics to me. Either it is obvious to the reader - in which case they don't need to be told that it's obvious - or it's not obvious to the reader, in which case you have left them with no help whatsoever and made yourself look arrogant (you may as well have called them stupid).

    To me, if something truly is obvious, then it is possible to give a comment that explains why. "If $\gcd(p,q) = 1$, we know that $p$ and $q$ share no prime factors and so neither do $p^m$ and $q^m$. Thus $\gcd(p^m, q^m) = 1$". Your comments don't need to be a complete proof, but should point the way for anyone who doesn't find your reasoning obvious. If you are writing an article for publishing, leave it for the editor to call his readers stupid should he so wish to edit your piece.
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  6. #6
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    Re: n-th root of natural number

    Here is another try...

    Suppose that $\sqrt[m]{n} \in \mathbb{Q}$ where $m, n \in \mathbb{N}.$

    Then $ \exists \ p,q \in \mathbb{Z}$ such that $ a>0, b > 0$ and $gcd(a, b) = 1$ where $\sqrt[m]{n} = \frac{a}{b}.$

    Hence $$ nb^m = a^m$$ and it follows that $b|a^m$.

    By the Fundamental Theorem of Arithmetic, $b$ can be written uniquely as a product of primes $\therefore \ \exists \ $ some prime number $p$ such that $p|b$. Hence $p|a^m$.

    Now by Euclid's Lemma we have that $p|a^m \implies p|a$ but since $gcd(a,b)=1$ there cannot be a prime $p$ such that $p|b$ and $p|a$ so we must have that $b=1$.

    Hence $n = a^m \in \mathbb{Z}$.

    Therefore if $\sqrt[m]{n} \in \mathbb{Q}$ then $n$ an integer $m^{th}$ power. Otherwise it is irrational.
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  7. #7
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    Re: n-th root of natural number

    I think you have a typo in line 3 and mean

    $\exists\ a,\ b \in \mathbb Z\ such\ that\ a > 0,\ b > 0,\ and\ gcd(a,\ b) = 1.\ where\ \sqrt[m]{n} = \dfrac{a}{b}.$

    But I think you could make it more concise as

    $\exists\ a,\ b \in \mathbb N\ such\ that\ gcd(a,\ b) = 1\ and\ \sqrt[m]{n} = \dfrac{a}{b}.$

    I don't think your line about the fundamental theorem of arithmetic is correct. I would say that by the fundamental theorem of arithmetic either b = 1, b is a prime, or b is a product of primes. Assume b is not equal to 1. Now you can say that there is a prime p that divides b and proceed as you did to show a contradiction. Therefore b = 1.

    I also think that you miss the boat with your line that $n = a^m \in \mathbb Z.$

    You started from the assumption that n was a natural number so it's hardly news that n is an integer.

    I think what is pertinent is

    $\sqrt[m]{n} = \dfrac{a}{b} = \dfrac{a}{1} = a \in \mathbb N.$

    So you have proved that $\sqrt[m]{n} \in \mathbb Q \implies \sqrt[m]{n} \in \mathbb N.$

    That is a stronger conclusion than you have provided.

    Finally, I have a question. Can you skip the steps where you say the root is real, that therefore the root is either rational or irrational, and that consequently the root is either an irrational number or a natural number? Those steps are logically necessary, but perhaps they are so obvious that they can be skipped.

    Those are my thoughts, but I am not a mathematician.
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